Simplify the following expression and state the condition under which the simplification is valid. You can assume that $a \neq 0$. $z = \dfrac{25a - 5}{3a} \times \dfrac{3a}{5a - 1} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ (25a - 5) \times 3a } { 3a \times (5a - 1) } $ $ z = \dfrac {3a \times 5(5a - 1)} {3a (5a - 1)} $ $ z = \dfrac{15a(5a - 1)}{3a(5a - 1)} $ We can cancel the $5a - 1$ so long as $5a - 1 \neq 0$ Therefore $a \neq \dfrac{1}{5}$ $z = \dfrac{15a \cancel{(5a - 1})}{3a \cancel{(5a - 1)}} = \dfrac{15a}{3a} = 5 $